A circuit consists of two very low-resistance rails separated by 0.300 m, with a 50.0 ohm resistor connected across them at one
1 answer:
Answer:
v = 288.67 m/s
Explanation:
Given that:
length of the separation I = 0.3
Resistance R = 50.0 ohms
Magnetic field B = 0.100 T
Power dissipated = 1.50 W
In a simple circuit, The Emf voltage of the power dissipated can be determined via the expression:
Power =
1.5 =
V = 8.660 V
From the Electromotive Force Emf ; we can calculate the speed of the bar
i.e Emf = B×v×l
8.660 = 0.1 × v × 0.3
v =
v = 288.67 m/s
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Answer:
a. It starts at point B.
vp = 2.53*10⁴ m/s
a. it starts at point A.
ve= 1.08*10⁶ m/s
Explanation:
a) As the proton is a positive charge, when released from rest, it will be accelerated due to the potential difference, from the higher potential to the lower one, so it is at the point B when released.
Once released, as the total energy must be conserved, the increase in kinetic energy must be equal (in magnitude) to the change in the electric potential energy, as follows:
ΔK + ΔUe = 0 ⇒ ΔK = -ΔUe =- (e*ΔV)
⇒
where e= elementary charge= 1.6*10⁻¹⁹ C, VA = 1.83 V, VB= 5.17V, and mp= mass of proton = 1.67*10⁻²⁷ kg.
Replacing by these values, and solving for v, we have:
⇒ vp = 2.53*10⁴ m/s
b) If, instead of a proton, the charge realeased from rest, had been an electron, a few things would change:
First, as the electrons carry negative charges, they move from the lower potentials to the higher ones, which means that it would have started at point A.
Second, as its charge is (-e) the change in electric potential energy had been negative also:
ΔUe = -e*ΔV = -e* (VB-VA)
In order to find the speed of the electron when it is just passing point B, we can apply the conservation of energy principle as for the proton, as follows:
where e= elementary charge= 1.6*10⁻¹⁹ C, VA = 1.83 V, VB= 5.17V, and me= mass of electron = 9.1*10⁻³¹ kg.
Replacing by these values, and solving for v, we have:
⇒ ve = 1.08*10⁶ m/s
Answer:
Explanation:
A differential equation that contain a term with di(t)/dt is in a RL circuit. Here we have
where vr is the voltage in the resistance, vi is the voltage in the inductance and vb is the source voltage. But also we have that
where L is the inductance of the circuit, r is the resistance an i is the current. By replacing we have the differential equation
I hope this is useful for you
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